SPOJ 7001. Visible Lattice Points 莫比乌斯反演
Description
Consider a N*N*N lattice. One corner is at (0,0,0) and the opposite one is at (N,N,N). How many lattice points are visible from corner at (0,0,0) ? A point X is visible from point Y iff no other lattice point lies on the segment joining X and Y.
Input :
The first line contains the number of test cases T. The next T lines contain an interger N
Output :
Output T lines, one corresponding to each test case.
Sample Input :
3
1
2
5
Sample Output :
7
19
175
Constraints :
T <= 50
1 <= N <= 1000000
Hint
| Added by: | Varun Jalan |
| Date: | 2010-07-29 |
| Time limit: | 1.368s |
| Source limit: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All except: NODEJS objc PERL 6 VB.net |
| Resource: | own problem used for Indian ICPC training camp |
题要求求满足gcd（a,b,c)=1 的个数 其中0<=a,b,c<=n
令f[n]=gcd(a,b,c)=n的个数
F[n]为gcd（a,b,c)=d的倍数的个数
F[n]=sigma（n|d，f（d））；即F[d]=（n/i）*（n/i)*(n/i)
f[n]=sigma(n|d，mu[d/n]*F[d]);
求f（1）=sigma（n，mu[i]*F[i]）;
因为这是从（0,0,0）开始所以F[d]=(n/i)*(n/i)*(n/i+3)///xyz上的点
ACcode：
#include <map>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define maxn 1000100
#define ll long long
using namespace std;
bool vis[maxn];
int prime[maxn];
int mu[maxn];
void get_M(){
memset(vis,false,sizeof(vis));
mu[1]=1;
int tot=0;
for(int i=2;i<=maxn;++i){
if(!vis[i]){
prime[tot++]=i;
mu[i]=-1;
}
for(int j=0;j<tot;j++){
if(i*prime[j]>maxn)break;
vis[i*prime[j]]=1;
if(i%prime[j]==0){
mu[i*prime[j]]=0;
break;
}
else mu[i*prime[j]]=-mu[i];
}
}
}
int main(){
int loop,n;
get_M();
scanf("%d",&loop);
while(loop--){
scanf("%d",&n);
ll ans=3;
for(int i=1;i<=n;++i)ans+=(long long)mu[i]*(n/i)*(n/i)*((n/i)+3);
printf("%lld\n",ans);
}
return 0;
}